In a finite lattice, every filter is principal.
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This proof feels to easy. Suggestions?
PROOF: We note that a filter $F$ is principal iff $land$$F$ $in$ $F$. (We define $land$$F$ = $land$${$$G$: $G$ $in$ $F$$}$) Suppose that $L$ is a finite lattice, and let $F$ be a filter on $L$. We aim to show $land$$F$ $in$ $F$. Since $L$ is finite, $L$ is complete, so that for every subset of $L$ meet and join are defined; hence by definition $land$$F$ exists. Hence $land$$F$ $in$ $F$, and F is principal, as claimed. END PROOF.
abstract-algebra proof-verification order-theory lattice-orders filters
edited Nov 24 at 3:43
Eric Wofsey
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asked Jan 11 '17 at 0:12
لويس العرب
38019
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up vote
4
down vote
favorite
This proof feels to easy. Suggestions?
PROOF: We note that a filter $F$ is principal iff $land$$F$ $in$ $F$. (We define $land$$F$ = $land$${$$G$: $G$ $in$ $F$$}$) Suppose that $L$ is a finite lattice, and let $F$ be a filter on $L$. We aim to show $land$$F$ $in$ $F$. Since $L$ is finite, $L$ is complete, so that for every subset of $L$ meet and join are defined; hence by definition $land$$F$ exists. Hence $land$$F$ $in$ $F$, and F is principal, as claimed. END PROOF.
abstract-algebra proof-verification order-theory lattice-orders filters
edited Nov 24 at 3:43
Eric Wofsey
178k12202330
asked Jan 11 '17 at 0:12
لويس العرب
38019
You can't just use completeness, because there are complete lattices where not all filters are principal.
– egreg
Jan 11 '17 at 0:23
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
This proof feels to easy. Suggestions?
PROOF: We note that a filter $F$ is principal iff $land$$F$ $in$ $F$. (We define $land$$F$ = $land$${$$G$: $G$ $in$ $F$$}$) Suppose that $L$ is a finite lattice, and let $F$ be a filter on $L$. We aim to show $land$$F$ $in$ $F$. Since $L$ is finite, $L$ is complete, so that for every subset of $L$ meet and join are defined; hence by definition $land$$F$ exists. Hence $land$$F$ $in$ $F$, and F is principal, as claimed. END PROOF.
abstract-algebra proof-verification order-theory lattice-orders filters
edited Nov 24 at 3:43
Eric Wofsey
178k12202330
asked Jan 11 '17 at 0:12
لويس العرب
38019
This proof feels to easy. Suggestions?
PROOF: We note that a filter $F$ is principal iff $land$$F$ $in$ $F$. (We define $land$$F$ = $land$${$$G$: $G$ $in$ $F$$}$) Suppose that $L$ is a finite lattice, and let $F$ be a filter on $L$. We aim to show $land$$F$ $in$ $F$. Since $L$ is finite, $L$ is complete, so that for every subset of $L$ meet and join are defined; hence by definition $land$$F$ exists. Hence $land$$F$ $in$ $F$, and F is principal, as claimed. END PROOF.
abstract-algebra proof-verification order-theory lattice-orders filters
abstract-algebra proof-verification order-theory lattice-orders filters
edited Nov 24 at 3:43
Eric Wofsey
178k12202330
asked Jan 11 '17 at 0:12
لويس العرب
38019
edited Nov 24 at 3:43
Eric Wofsey
178k12202330
asked Jan 11 '17 at 0:12
لويس العرب
38019
edited Nov 24 at 3:43
Eric Wofsey
178k12202330
edited Nov 24 at 3:43
Eric Wofsey
178k12202330
edited Nov 24 at 3:43
Eric Wofsey
178k12202330
178k12202330
asked Jan 11 '17 at 0:12
لويس العرب
38019
asked Jan 11 '17 at 0:12
لويس العرب
38019
asked Jan 11 '17 at 0:12
لويس العرب
38019
38019
You can't just use completeness, because there are complete lattices where not all filters are principal.
– egreg
Jan 11 '17 at 0:23
add a comment |
You can't just use completeness, because there are complete lattices where not all filters are principal.
– egreg
Jan 11 '17 at 0:23
You can't just use completeness, because there are complete lattices where not all filters are principal.
– egreg
Jan 11 '17 at 0:23
You can't just use completeness, because there are complete lattices where not all filters are principal.
– egreg
Jan 11 '17 at 0:23
add a comment |
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This is almost correct, but you haven't explained why $bigwedge Fin F$. Just because $bigwedge F$ exists in $L$ does not automatically mean it must be an element of $F$. You need to again use the fact that $F$ is finite, so $bigwedge F$ can be constructed by repeatedly taking binary meets using the elements of $F$. Since $F$ is closed under binary meets, this gives that $bigwedge Fin F$.
answered Jan 11 '17 at 0:19
Eric Wofsey
178k12202330
add a comment |
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up vote
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This is almost correct, but you haven't explained why $bigwedge Fin F$. Just because $bigwedge F$ exists in $L$ does not automatically mean it must be an element of $F$. You need to again use the fact that $F$ is finite, so $bigwedge F$ can be constructed by repeatedly taking binary meets using the elements of $F$. Since $F$ is closed under binary meets, this gives that $bigwedge Fin F$.
answered Jan 11 '17 at 0:19
Eric Wofsey
178k12202330
add a comment |
up vote
8
down vote
This is almost correct, but you haven't explained why $bigwedge Fin F$. Just because $bigwedge F$ exists in $L$ does not automatically mean it must be an element of $F$. You need to again use the fact that $F$ is finite, so $bigwedge F$ can be constructed by repeatedly taking binary meets using the elements of $F$. Since $F$ is closed under binary meets, this gives that $bigwedge Fin F$.
answered Jan 11 '17 at 0:19
Eric Wofsey
178k12202330
add a comment |
up vote
8
down vote
up vote
8
down vote
This is almost correct, but you haven't explained why $bigwedge Fin F$. Just because $bigwedge F$ exists in $L$ does not automatically mean it must be an element of $F$. You need to again use the fact that $F$ is finite, so $bigwedge F$ can be constructed by repeatedly taking binary meets using the elements of $F$. Since $F$ is closed under binary meets, this gives that $bigwedge Fin F$.
answered Jan 11 '17 at 0:19
Eric Wofsey
178k12202330
This is almost correct, but you haven't explained why $bigwedge Fin F$. Just because $bigwedge F$ exists in $L$ does not automatically mean it must be an element of $F$. You need to again use the fact that $F$ is finite, so $bigwedge F$ can be constructed by repeatedly taking binary meets using the elements of $F$. Since $F$ is closed under binary meets, this gives that $bigwedge Fin F$.
answered Jan 11 '17 at 0:19
Eric Wofsey
178k12202330
answered Jan 11 '17 at 0:19
Eric Wofsey
178k12202330
answered Jan 11 '17 at 0:19
Eric Wofsey
178k12202330
answered Jan 11 '17 at 0:19
Eric Wofsey
178k12202330
178k12202330
add a comment |
add a comment |
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You can't just use completeness, because there are complete lattices where not all filters are principal.
– egreg
Jan 11 '17 at 0:23